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Proof step 6 from proof put-at (previous step)

∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ 0 = n, (a ' + x0)[n] = (a0 ' + x0)[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1]
Proof step 8:
a ' + x0 ≠ [] ∧ n ≠ 0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ 0 = n, (a ' + x0).rest[n - 1] = (a0 ' + x0)[n]

Rule argument

spec: list-del inst: name: get-rec seq: ⊦ x ≠ [] ∧ n ≠ 0 → x[n] = x.rest[n - 1] subst: [n, x] → [n, a ' + x0] flags: rewrite, current

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, a ' + x0